Question

Из всех прямоугольников с периметром p найти прямоугольник наибольшей площади .

Answer 1

Дан прямоугольник со сторонами а и в
тогда периметр
$\displaystyle p=2(a+b)\\\\a+b=p/2\\\\a=p/2-b= \frac{p-2b}{2}$

найдем площадь

$\displaystyle S=a*b= \frac{p-2b}{2}*b= \frac{pb-2b^2}{2}= \frac{1}{2}(-2b^2+pb)$

при каком значении b площадь будет наибольшей? 
мы видим то что y=-2b²+pb это парабола, ветви вниз 
значит наибольшее значение она примет в своей вершине

$\displaystyle b_0= \frac{-p}{2*(-2)}= \frac{p}{4}$

значит если b=p/4 то площадь будет наибольшей. 
чему равно а? 
$\displaystyle a= \frac{p-2*p/4}{2}= \frac{p-p/2}{2}= \frac{p}{4}$

Значит если a=b=p/4 мы получим наибольшую площадь
и фигура у коротой а=в - квадрат

Answer 2

Finding the Rectangle with the Largest Area among Rectangles with a Given Perimeter

To find the rectangle with the largest area among rectangles with a given perimeter, we need to determine the dimensions of the rectangle that maximize its area. Let's denote the length of the rectangle as L and the width as W.

The perimeter of a rectangle is given by the formula: p = 2L + 2W.

To find the rectangle with the largest area, we need to maximize the area function, which is given by: A = L * W.

To solve this problem, we can use calculus. We need to express the area function in terms of a single variable, either L or W, and then find the critical points where the derivative of the area function is equal to zero.

Let's express the area function in terms of L: A = L * (p/2 - L).

To find the critical points, we differentiate the area function with respect to L and set it equal to zero: dA/dL = p/2 - 2L = 0.

Solving this equation for L, we get: L = p/4.

Substituting this value of L back into the perimeter equation, we can find the corresponding value of W: p = 2L + 2W, p = 2(p/4) + 2W, p/2 = p/2 + 2W, W = p/4.

Therefore, the rectangle with the largest area among rectangles with a given perimeter has dimensions L = p/4 and W = p/4.

Let's summarize the solution: - To find the rectangle with the largest area among rectangles with a given perimeter p, the dimensions of the rectangle should be L = p/4 and W = p/4.

Please note that the above solution assumes that the rectangle is a regular rectangle (all angles are right angles). If the problem allows for irregular rectangles, the solution may differ.

I hope this helps! Let me know if you have any further questions.